Understanding how Either is an instance of Functor

Question

In my free time I\'m learning Haskell, so this is a beginner question.

In my readings I came across an example illustrating how Either a is made an instance

Answer 1

As others mentioned, Either type is a functor in its both arguments. But in Haskell we are able to (directly) define only functors in a type's last arguments. In cases like this, we can get around the limitation by using newtypes:

newtype FlipEither b a = FlipEither { unFlipEither :: Either a b }

So we have constructor FlipEither :: Either a b -> FlipEither b a that wraps an Either into our newtype with swapped type arguments. And we have dectructor unFlipEither :: FlipEither b a -> Either a b that unwraps it back. Now we can define a functor instance in FlipEither's last argument, which is actually Either's first argument:

instance Functor (FlipEither b) where
    fmap f (FlipEither (Left x))  = FlipEither (Left (f x))
    fmap f (FlipEither (Right x)) = FlipEither (Right x)

Notice that if we forget FlipEither for a while we get just the definition of Functor for Either, just with Left/Right swapped. And now, whenever we need a Functor instance in Either's first type argument, we can wrap the value into FlipEither and unwrap it afterward. For example:

fmapE2 :: (a -> b) -> Either a c -> Either b c
fmapE2 f = unFlipEither . fmap f . FlipEither

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Update: Have a look at Data.Bifunctor, of which Either and (,) are instances of. Each bifunctor has two arguments and is a functor in each of them. This is reflected in Bifunctor's methods first and second.

The definition of Bifunctor of Either is very symetric:

instance Bifunctor Either where
    bimap f _ (Left a)  = Left (f a)
    bimap _ g (Right b) = Right (g b)

    first  f = bimap f id

    second f = bimap id f